The total mass flux (flow out of the box) is

1.72, Groundwat er Hydrology Prof. Charles Harvey

Le ct u r e Pa ck e t # 4 : Con t inu it y a n d Flow N e t s

Equa t ion of Con t inu it y

• Our equat ions of hydrogeology are a com binat ion of

o Conservat ion of m ass

o Som e em pir ical law ( Darcy’s Law , Fick’s Law )

• Develop a cont rol volum e, rect angular parallelepiped, REV ( Represent at ive Elem ent ary Volum e)

z

y

x dx

dy dz

x, y, z

q

m ass inflow rat e – m ass out flow rat e = change in m ass st orage qx = specific discharge in x- direct ion ( volum e flux per area)

at a point x,y,z L3/ L2- T

Consider m ass flow t hrough plane y- z at ( x,y,z)

x ρ dy dz L/ T M/ L3 L L = M/ T

Rat e of change of m ass flux in t he x- direct ion per unit t im e per cross- sect ion is

∂

]

[

ρ

q

dydz

dx

dy dz

x, y, z x

∂

x

m ass flow int o t he ent ry plane y- z is

∂

dx

]

[

ρ

q

x

dydz

−

[

ρ

q

x

]

dydz

∂

x

2

And m ass flow out of t he exit plane y- z is

∂

dx

]

[

ρ

q

x

dydz

+

[

ρ

q

x

]

dydz

∂

x

2

I n t he x- direct ion, t he flow in m inus t he flow out is

∂

]

−

[

ρ

q

_x

dxdydz

∂

x

Sim ilar ly, t he flow in t he y- direct ion t hrough t he plane dxdz ( figure on left )

dy dz

x, y, z

dx

dy dz

x, y, z

dx

⎡

⎤ ⎡

⎤

∂

⎢

[

ρ

q

dxdz

−

∂

[

ρ

q

_y

]

dy

dxdz

_{⎥ − ⎢}

_[

ρ

q

dxdz

+

∂

[

ρ

q

_y

]

dy

dxdz

_⎥

=

−

∂

y

[

ρ

q

dxdydz

⎣

y

]

∂

y

2

_{⎦ ⎣}

y

]

∂

y

2

_⎦

y

]

for t he net y- m ass flux.

Sim ilarly, we get for t he net m ass flux in t he z- direct ion:

∂

]

−

[

ρ

q

_z

dxdydz

∂

z

The t ot a l m a ss flu x ( flow out of t he box) is

⎡ ∂

[

ρ

q

]

∂

[

ρ

q

y

]

₋

∂

[

ρ

q

]

⎤

⎥

dxdydz

x

−

z

⎢−

x

St eady- st at e flow equat ion for het erogeneous, anisot ropic condit ions:

⎡

_⎞

⎤

For isot ropic, hom ogeneous condit ions ( K is not direct ional)

0

Note that the full equation at this point can be written in summation notation as:

ij

)

0

0

Flow N e t s

As w e have seen, t o w ork w it h t he groundw at er flow equat ion in any m eaningful w ay, w e have t o find som e kind of a solut ion t o t he equat ion. This solut ion is based on boundary condit ions, and in t he t ransient case, on init ial condit ions.

Let us look at t he t w o- dim ensional, st eady- st at e case. I n ot her words, let t he follow ing equat ion apply:

∂

⎛

∂

x

_⎝

⎜⎜

K

∂

∂

h

y

y

⎞

∂

∂ ⎞

⎟ +

∂

h

x

⎛

⎜

=

∂

x

⎝

K

x

⎠

⎟⎟

_⎠

( Map View)

A solut ion t o t his equat ion requires us t o specify boundary condit ion. For our purposes w it h flow net s, let us consider

∂

∂

boundary) .

• Const ant - head boundaries (h = const ant )

• Wat er- t able boundary ( free surface, h is not a const ant )

A relat ively st raight forward graphical t echnique can be used t o find t he solut ion t o t he GW flow equat ion for m any such sit uat ions. This t echnique involves t he

h

n

const ruct ion of a flow n e t.

A flow net is t he set of equipot ent ial lines ( const ant head) and t he associat ed flow lines ( lines along w hich groundw at er m oves) for a part icular set of boundary condit ions.

• For a given GW flow equat ion and a given value of K, t he boundary condit ions com plet ely det erm ine t he solut ion, and t herefore a flow net .

I n addit ion, let us first consider only hom ogeneous, isot ropic condit ions:

0

No- flow boundaries (

=

, where n is t he direct ion perpendicular t o t he

•

∂

2

h

∂

2

h

=

+

( Cross- Sect ion)

x

∂

2

z

C D

gr a ve l h1

E

1

5 2 4

3 7 8

h

h2

6 Sa n d a n d

I m pe r m e a ble La ye r

D a m

F

O G

A B

x y

h h1

C D

G

E F

B A

M y

h2

C h = 0

A h = h

h = h h2

H

B 6 0 ’

5 0 ’ 1 6 ’

G

q= q q5

q4

q3

q2

q1

q= 0 D

Let ’s look at flow in t he vicinit y of each of t hese boundaries. ( I sot ropic, hom ogeneous condit ions) .

N o- Flow Bou n da r ie s:

∂

= 0

∂

h

x

or

∂

_{= 0}

∂

h

y

or

∂

_{= 0}

∂

h

n

• Flow is parallel t o t he boundary.

• Equipot ent ials are perpendicular t o t he boundary Con st a n t - H e a d Bou nda r ie s: h = const ant

• Flow is perpendicular t o t he boundary.

• Equipot ent ials are parallel t o t he boundary. W a t e r Ta ble Bou n da r ie s: h= z

Anyw here in an aquifer, t ot al head is pressure head plus elevat ion head: h = ψ + z

However, at t he wat er t able, ψ = 0. Therefore, h = z

Ru le s for Flow N e t s ( I sot r opic, H om oge ne ou s Syst e m ) :

I n addit ion t o t he boundary condit ions t he follow ing rules m ust apply in a flow net : 1) Flow is perpendicular t o equipot ent ials everywhere.

2) Flow lines never int ersect .

3) The areas bet ween flow lines and equipot ent ials are “ curvilinear squares” . I n ot her words, t he cent ral dim ensions of t he “ squares” are t he sam e ( but t he flow lines or equipot ent ials can curve) .

• I f you draw a circle inside t he curvilinear square, it is t angent ial t o all four sides at som e point .

1

2 D a m

h= h

h= h Re se r voir

Why are t hese circles? I t preserves dQ along any st ream t ube. dQ = K dm ; dh/ ds = K dh

ds dQ

dQ

dQ dm

I f dm = ds ( i.e. ellipse, not circle) , t hen a const ant fact or is used. Ot h e r point s:

I t is not necessary t hat flow net s have finit e boundaries on all sides; r egions of flow t hat ext end t o infinit y in one or m ore direct ions are possible ( e.g., see t he figure above) .

A flow net can have “ part ial” st ream t ubes along t he edge. A flow net can have part ial squares at t he edges or ends of t he flow syst em .

Ca lcu la t ion s fr om Flow N e t s:

Probably t he m ost im port ant calculat ion is discharge from t he syst em . For a syst em wit h one recharge area and one discharge area, we can calculat e t he discharge wit h t he following expression:

Q = nfK dH H = nd dH Gives: Q = nf/ nd KH

Where Q is t he volum e discharge rat e per unit t hickness of sect ion perpendicular t o t he flow net ; nf is t he num ber of st ream t ubes ( or flow channels) ; nd is t he num ber

For hom ogeneous K,

∂

2

I nt roduce t he t ransform ed variable

Applying t his variable gives:

Kx is t he hydraulic conduct ivit y horizont ally on your page, and Kz is t he hydraulic conduct ivit y vert ically on your page. This t ransform at ion is not specific t o t he x-dim ension or t he y- x-dim ension.

2. On t he t ransform ed syst em , follow t he exact sam e principles for flow net s as out lined for a hom ogeneous, isot ropic syst em .

3. Perform t he inverse t ransform on t he syst em , i.e.

K

_z

Z

=

Z

'

K

_x

4. I f any flow calculat ions are needed, do t hese calculat ions on t he

hom ogeneous ( st ep 2) sect ion. Use t he follow ing for hydraulic conduct ivit y:

K

'

=

K

x

K

z

Where K’ is t he hom ogeneous hydraulic conduct ivit y of t he t ransform ed sect ion. ( NOTE: This t ransform ed K’ is not real! I t is only used for calculat ions on t he t ransform ed sect ion.)

Ex a m ple s:

T I

h = 0

1 layer syst em , you w ill only have curvilinear squares in one of t he layers. Which layer t o draw squares in is your choice: in general you should choose t he t hicker/ larger layer.

2

You can rearrange t he t angent law in any way t o det erm ine one unknown quant it y. For exam ple, t o det erm ine t he angle θ2:

⎛

⎜⎜

K

_K

⎞

θ

2

=

tan

−1

tan

_⎟⎟

1

⎠

2

θ

1

⎝

One im port ant consequence for a m edium wit h large cont rast s in K: high- K layers w ill oft en have alm ost horizont al flow ( in general) , w hile low - K layers w ill oft en have alm ost vert ical flow ( in general) .

Ex a m ple :

I n a t hree- layer syst em , K1 = 1 x 10- 3 m / s and K2 = 1 x 10- 4 m / s. K3 = K1. Flow in t he syst em is 14o _{below horizont al. What do flow in layers 2 and 3 look like?}

K2

K3 K1 76o

-4

θ

=

tan

−1

⎛

_⎜⎜

1x10

tan

76

o

_⎟⎟

⎞

=

22

o

⎝

1x10

-3

⎠

When draw ing flow net s w it h different layers, a very helpful quest ion t o ask is “ What layer allows wat er t o go from t he ent rance point t o t he exit point t he easiest ?” Or, in ot her words, “ What is t he easiest ( frict ionally speaking) way for wat er t o go from here t o t here?”

K2 K1

K1 K2

K2 K1

K

K

1

=

10